Thermal Resistance due to Convection Heat Transfer

1) A typical wall construction for a single-family residence might consist of 0.375 inches of outer sheathing (insulating board, hardwood siding, etc.), 3.5 inches of mineral fiber insulation and then a 0.375 inches inner surface (gypsum, etc.). Typical thermal conductivities for these materials are 0.1 W/m-C for the siding, 0.046 W/m-C for the insulation and 0.17 W/m-C for the inner surface. Determine the total thermal resistance (hr-ft2-F/Btu) and total wall conductance (Btu/hr-ft2-F). Also find the percent of total resistance for the three different materials. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, effects of each wall-layer on heat flow, etc (2.5 points).

2) Part 1 neglected thermal resistances due to convection heat transfer on the inner and outer wall surfaces. Recalculate the total thermal resistance (along with percent) and the wall U-factor including convection heat transfer using an outer wall convection heat transfer coefficient of 6.0 Btu/hr-ft2-F (15 mph wind in the winter) and an inner wall value of 1.46 Btu/hr-ft2-F. Provide step-by-step calculation procedures. Discuss/comment to include as a minimum things like assumptions, comparisons, etc (2.5 points).

Note: Both the siding, and the inner wall surface have provide very little R-value. The insulation allows for thinner walls, as it would take an abundant amount of either siding or inner surface to achieve an acceptable level of thermal resistance.

Wall Conductance (C)

C=1/R=1/137 Btu/(hr×ft^2×℉)

Note: Wall conductance is simply the inverse of the R-value, often expressed as U-value. It measures the rate of heat flow though a unit area of a material. This should not be confused with thermal conductivity (k), which is an inherent property of the material.

Part 2) I have been staring at this for hours trying to understand what I'm not understanding! I know it must have something to do with the heat loss on the inside being equal to the heat gain on the outside, but I just cannot figure out how to get started on this problem without knowing the temperature difference.

Regarding part 2): You don't need a temperature to calculate the resistance (or the u-value), you only would need it if you want to calculate the heat passing the wall. So you simply can calculate the total resistance/u-value with the given heat transfer coefficients and the thermal conductivities.

Regarding part1): I didn't check it, because it seemed to be finished. Need something special with that?

Basically the formula is correct, but just to be sure about the (1/ktotal)-part. In your first post you defined k as the heat conductivity in W / (m⋅K). If ktotal has the same dimension, the equation isn't finished.